Question: Complete the square to solve for $x$. $x^{2}+3x-28 = 0$
Explanation: Move the constant term to the right side of the equation. $x^2 + 3x = 28$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. The coefficient of our $x$ term is $3$ , so half of it would be $\dfrac{3}{2}$ , and squaring it gives us ${\dfrac{9}{4}}$ $x^2 + 3x { + \dfrac{9}{4}} = 28 { + \dfrac{9}{4}}$ We can now rewrite the left side of the equation as a squared term. $( x + \dfrac{3}{2} )^2 = \dfrac{121}{4}$ Take the square root of both sides. $x + \dfrac{3}{2} = \pm\dfrac{11}{2}$ Isolate $x$ to find the solution(s). $x = -\dfrac{3}{2}\pm\dfrac{11}{2}$ The solutions are: $x = 4 \text{ or } x = -7$ We already found the completed square: $( x + \dfrac{3}{2} )^2 = \dfrac{121}{4}$